The relation $3t = \sqrt{3x} + 6$ describes the displacement of a particle in one direction where $x$ is in me

Question

The relation $3t = \sqrt{3x} + 6$ describes the displacement of a particle in one direction where $x$ is in metres and $t$ in seconds. The displacement, when velocity is zero, is:

Accepted Answer

1.  **Rearrange the Equation:** Isolate $x$ in the given relation $3t = \sqrt{3x} + 6$.
    $\sqrt{3x} = 3t - 6$
    Squaring both sides:
    $3x = (3t - 6)^2$
    $3x = 9(t - 2)^2$
    $x = 3(t - 2)^2$
2.  **Find Velocity ($v$):** Instantaneous velocity is the rate of change of displacement with respect to time, $v = \frac{dx}{dt}$ .
    Differentiating $x$ with respect to $t$:
    $v = \frac{d}{dt}[3(t - 2)^2]$
    $v = 3 \cdot 2(t - 2) \cdot 1$
    $v = 6(t - 2)$.
3.  **Condition for Zero Velocity:** Set $v = 0$.
    $6(t - 2) = 0 \implies t = 2 	ext{ s}$.
4.  **Calculate Displacement:** Find the displacement $x$ at $t = 2 	ext{ s}$.
    Substitute $t = 2$ into the position equation:
    $x = 3(2 - 2)^2$
    $x = 3(0)^2 = 0 	ext{ metres}$.

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